Hi! Most of my Google quires lead to this forum so I figured I join! Seams like a good place for this kind of stuff.

Recently got in to DIY:ing some modules for my modular synth (mostly Modcan A). Most of the projects are based around logic chips (4000 series) and PIC:s. What I'm wondering is, before I let these modules play with my other modules, what precautions should I take for the input / output. Most circuits I come across has either or both a buffer before the output (usually an op-amp) and a protective resistor (usually 10-100k). Is this considered common practice?

When reading up on the CMOS logic chips, I came across this document discussing buffered (suffix B) and unbuffered (suffix UB) logic ICs. I'm curious to know if this would eliminate the need for a buffer afterwards.

Bonus question: If the last stage of a circuit is an op-amp (a mixer for example), it doesn't make any sense buffering that output right?

I usually use 1k on outputs, following the example of a lot of the stuff I've built by Thomas Henry. Can't say so much about inputs, Most of the stuff I've thrown together myself has been CV generators only (from arduino and such).

I can't say I know much about Buffered/Unbuffered CMOS, so I'll defer that to someone else.

I agree, buffering a mixer output doesn't seem like it should be necessary.

It's bad practice to rely on the internal buffers because they can only source limited current. This is fine for CMOS-to-CMOS connections, but in a modular synth context you can't be sure where the output will go. So generally we re-buffer with an opamp, and then after that a current-limiting resistor (1K generally) to protect against shorts.

At the inputs, there is a different problem: logic levels. At 12 volts, CMOS will switch to logic high at around 8.4 volts (70%), so this won't switch at all from a 5V gate. Commonly, transistors are used here to boost the input signal. Ideally you want it to switch at a pretty low level for easy compatibility with a variety of signal sources... perhaps 1.2 to 2.5 volts.

Another problem at the inputs is that CMOS cannot take any voltage outside of its supply rails, and its supply is single-sided -- on a bipolar +/-12V supply you normally give the CMOS subcircuit 12V and GND. Therefore, any negative voltage reaching the CMOS input pin will be routed through the chip's power protection network, and those tiny diodes can't handle much. You will burn up the chip. Therefore, it must also be protected against negative voltages. Typically this is solved with an inline diode.

Thanks a lot for the info! Very informative. You just saved a few small diodes from burning. They will be grateful.

Just out of curiosity.. I came across this CGS diagram while searching for info. In the upper right corner, there's something that looks like a transistor buffer(?). Most transistor buffers I've seen takes a Vref which is added to the base. Would something like that be enough for the outs of a CD4xxxxB? I have a lot of transistors laying around and since I'm only dealing with CVs right now (0-5v) I can live without bipolarity.

Yeah that's a buffer. I'm not sure what the trade-offs are between a transistor and an opamp in this application. No DC offset should be added, though (no VRef).

Edit, from an EE friend: "Either work fine. For logic transistors are way easier to use. For analog scaling op amps are easier. Also keep in mind that that circuit [CGS] can drive a crapload of current just by changing the transistor and adding cooling."

Cool, thanks! Seems like every time I think I got the hang of something, a ton of other questions pop-up. Still a long way to go

Anyway, this is what I've come up with so far. The first stage amplifies the signal and then passes in into a Schmitt trigger. Most input will be CV but I do want the option to pass in AC. Any remarks / suggestions are of course appreciated.

Some circuits (I'm thinking of a few different Thomas Henry circuits I've seen) put a switchable cap across the input to let you limit to AC only. Open switch, the signal goes through the cap and is AC coupled, closed switch, the cap is shorted and the signal is DC coupled.

Right, that was what I thought you meant when mentioning AC. Overall, yes, you don't have to filter it, assuming you're not going to be sending in an AC signal of say, 10vpp at a DC offset of 5+, it ought to be fine.

Edit: not even 10vpp at 5V is really a problem, but I get the point across too.

ah, surprised I didn't find that circuit before. Looks pretty much like what I want to do (though the purpose of this build is more to learn than anything).

Speaking of part count. I guess one alternative would be to just use a regulator and operate the whole circuit in the 0-5v range. Then use 5v zener diode at the input to protect against anything above 5v or negative (using the configuration in the diagram you posted but with a zener). This would eliminate the need for boosting the signal, leaving just the schmitt trigger at the input stage.. if the schmitt trigger has a high enough input impedance that is. More interestingly, the output buffers could then just be kept to a op-amp buffer without the need for any scaling.

I'm probably missing something, but wouldn't the schottky to ground take care of that? I.e anything above the breakdown voltage will go to ground. Does schottky diodes work differently from zener in other ways than the switching time?

Yes, a zener has a so called 'zener voltage', a voltage applied in reverse when the zener begins to conduct. Where a normal diode won't conduct in reverse, a zener will above that voltage.
A schottky doesn't have that property, but has a lower forward voltage drop. Where a normal silicon diode has a forward voltage drop of around 0.7V, for a schottky that's about 0.2V, allowing it to switch faster._________________Weblog!

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